The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
A barometer is used to measure air pressure!
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
Answer:
There are two reasons why air pressure decreases as altitude increases: density and depth of the atmosphere. Most gas molecules in the atmosphere are pulled close to Earth's surface by gravity, so gas particles are denser near the surface.
Explanation: